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Q25

Expert-verifiedFound in: Page 176

Book edition
Student Edition

Author(s)
Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

Pages
227 pages

ISBN
9780395977279

Write a paragraph proof.

Given: $\u25b1ABCD;\text{\hspace{0.17em}\hspace{0.17em}}\u25b1BEDF$

Prove: *AECF* is a $\u25b1$.

(Hint: A short proof is possible if certain auxiliary segments ate drawn.)

Two pairs of opposite sides $\overrightarrow{AE}\cong \overrightarrow{FC}$ and $\overrightarrow{AF}\cong \overrightarrow{CE}$in *AECF* are parallel therefore, *AECF* is a parallelogram.

In triangles $\Delta DAE$ and $\Delta BCF$.

$\overrightarrow{AD}\cong \overrightarrow{BC\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}\therefore $ (*ABCD* is a rhombus)

$\angle ADC\cong \angle ABC\therefore $ (*ABCD* is a rhombus)

$\angle ADE\cong \angle CBF\text{\hspace{0.17em}\hspace{0.17em}}\therefore $ ( $\angle EDC\&\angle ABF$ are the equal angle of parallelogram made of parallel sides two parallelograms)

*ABCD* is a parallelogram.

Similarly, $\angle ADC\cong \angle ABC$ ( *ABCD* is a rhombus),

$\angle EDC\&\angle ABF$ are equal angles of parallelogram made of parallel sides.

$\angle ADE\cong \angle CBF$

$\overrightarrow{DE}\cong \overrightarrow{BF}$ $\left(\therefore BEDF\text{\hspace{0.17em}is a\hspace{0.17em}parallelogram}.\right)$

Therefore, by SAS postulate $\Delta DAE\cong \Delta BCF$.

By CPCT, $\overrightarrow{AE}\cong \overrightarrow{FC}$.

In triangles $\Delta AFB\&\Delta CDE$.

$\overrightarrow{DE}\cong \overrightarrow{BF}\therefore \left(BEDF\text{\hspace{0.17em}is a parallelogram}\right)$

Since, $\angle EDC\&\angle ABF$ are equal angles of parallelogram made of parallel sides.

$\angle ADE\cong \angle CBF\text{\hspace{0.17em}\hspace{0.17em}}$

$\overrightarrow{DC}\cong \overrightarrow{BA}\therefore \left(BEDF\text{\hspace{0.17em}is a parallleogram}\right)$

Therefore, by SAS postulate $\Delta AFB\cong \Delta CED$.

By CPCT, $\overrightarrow{AF}\cong \overrightarrow{CE}$.

Since two pairs of opposite sides $\overrightarrow{AE}\cong \overrightarrow{FC}$ and $\overrightarrow{AF}\cong \overrightarrow{CE}$ in *AECF* are parallel, therefore, *AECF* is a parallelogram.

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